Simplifying $\prod_{i=1}^p(1-m\lambda_i)$, where $m$ is a constant and the $\lambda_i$ are the eigenvalues of a positive-definite symmetric matrix I have the following expression that I would like to simplify:
$$\prod_{i=1}^p\left( 1-m\lambda_i\right)$$
with $m := \dfrac{n-1}{n\sigma^2}$, where $n\in\mathbb{N}$, $\sigma^2>0$, and the $\lambda_i$ represent the $p$ eigenvalues of a positive-definite and symmetric matrix.
I'm sure, there is a formula somewhere that can be used to expand or approximate the above expression, but I don't see it... The expansion of the product gives something like $1+\dots$, which is what I'm looking for my purpose.
Any suggestion?
 A: My conjecture is that:
$$\displaystyle \prod_{i=1}^p (1-m \lambda_i)=1+\sum_{j=1}^{p-1}(-1)^j m^j\sum_{1\leq i_1<\ldots<i_j\leq p} \lambda_{i_1}\cdots \lambda_{i_j} +(-1)^p m^p \lambda_1\cdots \lambda_p.$$
Maybe you can check this via induction? Does that help? 
Let us try some low order cases:


*

*for $p=2$:


\begin{align*}
\textrm{RHS}&= 1+\sum_{j=1}^1 (-1)^j m^j \sum_{1\leq i_1<\ldots<i_j\leq 2} \lambda_{i_1}\cdots \lambda_{i_j} +m^2 \lambda_1 \lambda_2 \\
&=1- m(\lambda_1+\lambda_2)+m^2 \lambda_1 \lambda_2.
\end{align*}
Now 
\begin{align*}
\textrm{LHS}&=(1-m\lambda_1)(1-m\lambda_2)\\
&=1-m\lambda_2 -m\lambda_1+m^2 \lambda_1 \lambda_2=1-m(\lambda_1+\lambda_2)+m^2 \lambda_1 \lambda_2.
\end{align*}


*

*for $p=3$:


\begin{align*}
\textrm{RHS}&=1+\sum_{j=1}^2 (-1)^j m^j \sum_{1\leq i_1<\ldots<i_j\leq 3}\lambda_{i_1}\cdots \lambda_{i_j}+m^3 \lambda_1\lambda_2 \lambda_3\\
&=1-m(\lambda_1+\lambda_2+\lambda_3)+m^2(\lambda_1 \lambda_2+\lambda_1 \lambda_3+\lambda_2\lambda_3)-m^3\lambda_1 \lambda_2\lambda_3. 
\end{align*}
On the other hand:
\begin{align*}
\textrm{LHS}&=(1-m\lambda_1)(1-m\lambda_2)(1-m\lambda_3)\\
&=(1-m\lambda_2-m\lambda_1+m^2 \lambda_1 \lambda_2)(1-m\lambda_3)\\
&=1-m\lambda_3-m\lambda_2+m^2\lambda_2 \lambda_3-m\lambda_1+m^2 \lambda_1 \lambda_3+m^2 \lambda_1 \lambda_3-m^3 \lambda_1 \lambda_2 \lambda_3\\
&=1-m(\lambda_1+\lambda_2+\lambda_3)+m^2 (\lambda_1 \lambda_2+\lambda_1 \lambda_3+\lambda_2\lambda_3)-m^3 \lambda_1 \lambda_2 \lambda_3.
\end{align*}
A: Not sure if this helps, but just in case: if $A$ is the positive definite matrix in question, then 
$$
\chi_A(z):=\det(zI-A)=\prod_{i=1}^p (z-\lambda_i),
$$
and so 
$$
\prod_{i=1}^p(1-m\lambda_i)=m^p\prod_{i=1}^p (1/m-\lambda_i)=m^p\chi_A(1/m).
$$
This recovers PtF's answer above. Computing determinants is relatively inexpensive, so at least computationally this gives a nice way to evaluate your expression, assuming you have access to $A$.
Edit: As mentioned above, this gives PtF's answer. It is well-known that
$$
\prod_{i=1}^p (z-\lambda_i)=\sum_{i=0}^p (-1)^ie_i(\lambda_1,\ldots,\lambda_p)z^{p-i},
$$
where $e_i$ gives the $i$th symmetric polynomial i.e.
$$
e_i(\lambda_1,\ldots,\lambda_p)=\sum_{S\subseteq [p]: \vert S\vert=i} \prod_{j\in S} \lambda_j.
$$
We then have
$$
\prod_{i=1}^p (1-m\lambda_i)=m^p\prod_{i=1}^p (1/m-\lambda_i)=m^p\sum_{i=0}^p (-1)^ie_i(\lambda_1,\ldots,\lambda_p)m^{i-p}=\sum_{i=0}^p (-1)^ie_i(\lambda_1,\ldots,\lambda_p)m^{i},
$$
as was claimed.
